Mean value theorem Mean Value Theorem and Taylor's Theorem L'Hopital's rule Taylor's theorem Theorem Let f be a function on (a; b) and ξ 2 (a; b) such that 1 f is differentiable at x = ξ. 2 Either f (x) ≤ f (ξ) for any x 2 (a; b), or f (x) ≥ f (ξ) for any x 2 (a; b). Then f 0(ξ) = 0. Proof. Suppose f (x) ≤ f (ξ) for any x 2 (a; b). The proof for the other case is more or less the same. For any h < 0 with a < ξ + h < ξ, we have f (ξ + h) − f (ξ) ≤ 0 and h is negative. Thus f (ξ + h) − f (ξ) f 0(ξ) = lim ≥ 0 h!0− h On the other hand, for any h > 0 with ξ < ξ + h < b, we have f (ξ + h) − f (ξ) ≤ 0 and h is positive. Thus we have f (ξ + h) − f (ξ) f 0(ξ) = lim ≤ 0 h!0+ h Therefore f 0(ξ) = 0. MATH1010 University Mathematics Mean value theorem Mean Value Theorem and Taylor's Theorem L'Hopital's rule Taylor's theorem Theorem (Rolle's theorem) Suppose f (x) is a function which satisfies the following conditions. 1 f (x) is continuous on [a; b]. 2 f (x) is differentiable on (a; b). 3 f (a) = f (b) Then there exists ξ 2 (a; b) such that f 0(ξ) = 0. Proof. By extreme value theorem, there exist a ≤ α; β ≤ b such that f (α) ≤ f (x) ≤ f (β) for any x 2 [a; b]: Since f (a) = f (b), at least one of α; β can be chosen in (a; b) and we let it be ξ. Then we have f 0(ξ) = 0 since f (x) attains its maximum or minimum at ξ. MATH1010 University Mathematics Mean value theorem Mean Value Theorem and Taylor's Theorem L'Hopital's rule Taylor's theorem Theorem (Lagrange's mean value theorem) Suppose f (x) is a function which satisfies the following conditions. 1 f (x) is continuous on [a; b]. 2 f (x) is differentiable on (a; b). Then there exists ξ 2 (a; b) such that f (b) − f (a) f 0(ξ) = : b − a Proof f (b) − f (a) Let g(x) = f (x) − (x − a). Since g(a) = g(b) = f (a), by Rolle's theorem, b − a there exists ξ 2 (a; b) such that g 0(ξ) = 0 f (b) − f (a) f 0(ξ) − = 0 b − a f (b) − f (a) f 0(ξ) = b − a MATH1010 University Mathematics Mean value theorem Mean Value Theorem and Taylor's Theorem L'Hopital's rule Taylor's theorem Exercise (True or False) Suppose f (x) is a function which is differentiable on (a; b). 1 f (x) is constant on (a; b) if and only if f 0(x) = 0 on (a; b). Answer: T 2 f (x) is monotonic increasing on (a; b) if and only if f 0(x) ≥ 0 on (a; b). Answer: T 3 If f (x) is strictly increasing on (a; b), then f 0(x) > 0 on (a; b). Answer: F 4 If f 0(x) > 0 on (a; b), then f (x) is strictly increasing on (a; b). Answer: T f 0(x) > 0 + Strictly increasing + Monotonic increasing , f 0(x) ≥ 0 MATH1010 University Mathematics Mean value theorem Mean Value Theorem and Taylor's Theorem L'Hopital's rule Taylor's theorem Theorem Let f (x) be a function which is differentiable on (a; b). Then f (x) is monotonic increasing if and only if f 0(x) ≥ 0 for any x 2 (a; b). Proof Suppose f (x) is monotonic increasing on (a; b). Then for any x 2 (a; b), we have f (x + h) − f (x) ≥ 0 for any h > 0 and thus f (x + h) − f (x) f 0(x) = lim ≥ 0: h!0+ h On the other hand, suppose f 0(x) ≥ 0 for any x 2 (a; b). Then for any α < β in (a; b), by Lagrange's mean value theorem, there exists ξ 2 (α; β) such that f (β) − f (α) = f 0(ξ)(β − α) ≥ 0: Therefore f (x) is monotonic increasing on (a; b). Corollary f (x) is constant on (a; b) if and only if f 0(x) = 0 for any x 2 (a; b). MATH1010 University Mathematics Mean value theorem Mean Value Theorem and Taylor's Theorem L'Hopital's rule Taylor's theorem Theorem If f (x) is a differentiable function such that f 0(x) > 0 for any x 2 (a; b), then f (x) is strictly increasing. Proof. Suppose f 0(x) > 0 for any x 2 (a; b). For any α < β in (a; b), by Lagrange's mean value theorem, there exists ξ 2 (α; β) such that f (β) − f (α) = f 0(ξ)(β − α) > 0: Therefore f (x) is strictly increasing on (a; b). The converse of the above theorem is false. Example f (x) = x 3 is strictly increasing on R but f 0(0) = 0 is not positive. MATH1010 University Mathematics Mean value theorem Mean Value Theorem and Taylor's Theorem L'Hopital's rule Taylor's theorem Theorem (Cauchy's mean value theorem) Suppose f (x) and g(x) are functions which satisfies the following conditions. 1 f (x); g(x) is continuous on [a; b]. 2 f (x); g(x) is differentiable on (a; b). 3 g 0(x) 6= 0 for any x 2 (a; b). Then there exists ξ 2 (a; b) such that f 0(ξ) f (b) − f (a) = : g 0(ξ) g(b) − g(a) Proof f (b) − f (a) Let h(x) = f (x) − (g(x) − g(a)). g(b) − g(a) Since h(a) = h(b) = f (a), by Rolle's theorem, there exists ξ 2 (a; b) such that f (b) − f (a) f 0(ξ) − g 0(ξ) = 0 g(b) − g(a) f 0(ξ) f (b) − f (a) = g 0(ξ) g(b) − g(a) MATH1010 University Mathematics Mean value theorem Mean Value Theorem and Taylor's Theorem L'Hopital's rule Taylor's theorem Theorem (L'Hopital's rule) Let a 2 [−∞; +1]. Suppose f and g are differentiable functions such that 1 lim f (x) = lim g(x) = 0 (or ±∞). x!a x!a 2 g 0(x) 6= 0 for any x 6= a (on a neighborhood of a). f 0(x) 3 lim = L. x!a g 0(x) f (x) f (x) Then the limit of at x = a exists and lim = L. g(x) x!a g(x) Proof For any x 6= 0, by Cauchy's mean value theorem, there exists ξ between a and x such that f 0(ξ) f (x) − f (a) f (x) = = : g 0(ξ) g(x) − g(a) g(x) Here we redefine f (a) = g(a) = 0, if necessary, so that f and g are continuous at a. Note that ξ ! a as x ! a. We have f (x) f 0(ξ) lim = lim = L: x!a g(x) x!a g 0(ξ) MATH1010 University Mathematics Mean value theorem Mean Value Theorem and Taylor's Theorem L'Hopital's rule Taylor's theorem 0 1 Example (Indeterminate form of types and ) 0 1 sin x − x cos x x sin x 1 1: lim = lim = x!0 x 3 x!0 3x 2 3 x 2 2x 2x 2 2: lim = lim = lim = lim = 2 x!0 x!0 sec x tan x x!0 x!0 2 ln sec x sec x tan x sec x 2 3 x 2 ln(1 + x ) 3 1 x 3: lim = lim 1+x = lim lim x!0 x − sin x x!0 1 − cos x x!0 1 + x 3 x!0 1 − cos x 2x = lim = 2 x!0 sin x 3 4 4x 3 2 ln(1 + x ) 4 4x (1 + x ) 4: lim = lim 1+x = lim = 2 x!+1 ln(1 + x 2) x!+1 2x x!+1 2x(1 + x 4) 1+x2 MATH1010 University Mathematics Mean value theorem Mean Value Theorem and Taylor's Theorem L'Hopital's rule Taylor's theorem Example (Indeterminate form of types 1 − 1 and 0 · 1) 1 1 x − 1 − ln x 1 − 1 5: lim − = lim = lim x x!1 ln x x − 1 x!1 (x − 1) ln x x!1 x−1 x + ln x x − 1 1 1 = lim = lim = x!1 x − 1 + x ln x x!1 2 + ln x 2 −1 1 tan x 2 6: lim cot 3x tan−1 x = lim = lim 1+x x!0 x!0 tan 3x x!0 3 sec2 3x 1 1 = lim = x!0 3(1 + x 2) sec2 3x 3 cos x ln sin x sin x 7: lim x ln sin x = lim 1 = lim 1 x!0+ x!0+ x!0+ x − x2 −x 2 cos x = lim = 0 x!0+ sin x x + 1 ln(x + 1) − ln(x − 1) 8: lim x ln = lim x!+1 x − 1 x!+1 1 x 1 − 1 2x 2 = lim x+1 x−1 = lim = 2 x!+1 1 x!+1 (x + 1)(x − 1) − x2 MATH1010 University Mathematics Mean value theorem Mean Value Theorem and Taylor's Theorem L'Hopital's rule Taylor's theorem Example (Indeterminate form of types 00, 11 and 10) Evaluate the following limits.